Intelligent Systems II · TEE · papers

Q7(a) Bayes bag posterior [5]

Final examination · 17 Apr 2024

Actual historical paper ·100 marks attempted /140 offered · Q1 compulsory + any four of Q2–Q7. Source defects and solution assumptions are explicitly identified.

Question. Q7(a) [5 marks · 2024-04-17]

Bag B₁ contains 3 white and 7 black balls. Bag B₂ contains 5 white and 3 black balls. A ball drawn from one of the bags is white. Find the probability it came from B₁.

Answer

Given

P(W|B₁)=3/10; P(W|B₂)=5/8.

To find

P(B₁|W).

Formula and why it applies

Bayes reverses the condition. Weight the likelihood for each bag by its prior chance of being chosen, then normalize over all ways to obtain a white ball.

P(B_1\mid W)=\frac{P(W\mid B_1)P(B_1)}{P(W\mid B_1)P(B_1)+P(W\mid B_2)P(B_2)}LaTeX: P(B_1\mid W)=\frac{P(W\mid B_1)P(B_1)}{P(W\mid B_1)P(B_1)+P(W\mid B_2)P(B_2)}

Solution

Step 1. Calculate both white-event contributions
P(W,B_1)=\frac3{10}\frac12=\frac3{20}=0.15LaTeX: P(W,B_1)=\frac3{10}\frac12=\frac3{20}=0.15
P(W,B_2)=\frac58\frac12=\frac5{16}=0.3125LaTeX: P(W,B_2)=\frac58\frac12=\frac5{16}=0.3125
Step 2. Find total white probability
P(W)=\frac3{20}+\frac5{16}=\frac{12+25}{80}=\frac{37}{80}=0.4625LaTeX: P(W)=\frac3{20}+\frac5{16}=\frac{12+25}{80}=\frac{37}{80}=0.4625
Step 3. Normalize the B₁ contribution
P(B_1\mid W)=\frac{3/20}{37/80}=\frac3{20}\frac{80}{37}=\frac{12}{37}\approx0.324324LaTeX: P(B_1\mid W)=\frac{3/20}{37/80}=\frac3{20}\frac{80}{37}=\frac{12}{37}\approx0.324324
Final answer. Under equal bag-selection priors, 12/37 ≈ 0.324324.